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CAT Question
by Darshit Paun - Monday, 23 July 2012, 08:50 PM
  Q4)Points S ,T and U are selected on sides QR, PR, and PQ respectively of ∆PQR. The lines PS, QT, and RU meet at point Z. If area(PUZ) = 126, area(UQZ) = 63, and area(RTZ) = 24, the area of ∆PQR is ?
(a) 324 (b) 351 (c) 360 (d) 364.

Please post the solution!
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Re: CAT Question
by rakshit surana - Monday, 23 July 2012, 10:13 PM
  ans 351
in PZU and QZU
[1/2 * PU * h ] /[ 1/2 * QU * h] = 126/63 = 2/1
PU:QU = 2:1
Let area of PZT = X ; QZS = Y ; RZS = Z
SO AREA OF RUP : RUQ = 2:1
X + 126 +24 = 2(63+ Y +Z)
X + 24 = 2(Y+Z) ---------------------(1)

NOW IN PZT AND RZT
PT: RT = X:24
AND IN PQT AND RQT
PT:RT = 189+X : 24+Y+Z
SO
X/24= 189+X/ 24+Y+Z -----------------(2)
ELIMINATING Y+Z FROM (1) AND(2)
X^2 + 24X - 48*189 = 0
(X+108)(X-84)=0
X= 84
SO AREA OF PUR = 234
SO AREA OF PQR = 3/2 *234 = 351...
TRICKY QUESTION BUT NYC smile
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Re: CAT Question
by Darshit Paun - Monday, 23 July 2012, 11:10 PM
  Okay Thanks. I was unable to understand how a concurrent point can divide the areas of the triangles into the same ratios!!...
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Re: CAT Question
by Rajat Soni - Tuesday, 24 July 2012, 07:15 PM
  HOW CAN U ASSUME THAT PZU AND QZU ARE RIGHT ANGLE TRIANGLE.THERE IS NOTHING MENTIONED IN QUESTION EXCEPT AREA.
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